Question: Solve for $r$, $ \dfrac{2}{16r} = -\dfrac{8}{20r} - \dfrac{4r - 1}{4r} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16r$ $20r$ and $4r$ The common denominator is $80r$ To get $80r$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{2}{16r} \times \dfrac{5}{5} = \dfrac{10}{80r} $ To get $80r$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{20r} \times \dfrac{4}{4} = -\dfrac{32}{80r} $ To get $80r$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ -\dfrac{4r - 1}{4r} \times \dfrac{20}{20} = -\dfrac{80r - 20}{80r} $ This give us: $ \dfrac{10}{80r} = -\dfrac{32}{80r} - \dfrac{80r - 20}{80r} $ If we multiply both sides of the equation by $80r$ , we get: $ 10 = -32 - 80r + 20$ $ 10 = -80r - 12$ $ 22 = -80r $ $ r = -\dfrac{11}{40}$